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3 years ago

If you horizontally stretch the quadratic parent function, f(x) = x2, by a factor

Mathematics

1 answer:

Brilliant_brown [7]3 years ago

6 0

Answer:

Step-by-step explanation:

If you horizontally stretch the quadratic parent function, f(x) = x2, by a factor

of 3, what is the equation of the new function?

A. g(x) = 1/3x2

B. g(x) = 3x2

C. g(x) = (3x)2

D. g(x) = (1/3x)2

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2 years ago

Which of the following represents the zeros of f(x) = 2x3 − 5x2 − 28x + 15?

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$\mathrm{Use\:the\:rational\:root\:theorem}$

$a_0=15,\:\quad a_n=2$

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$\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:3,\:5,\:15}{1,\:2}$

$-\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+3$

$\mathrm{Compute\:}\frac{2x^3-5x^2-28x+15}{x+3}\mathrm{\:to\:get\:the\:rest\:of\:the\:eqution:\quad }2x^2-11x+5$

$=\left(x+3\right)\left(2x^2-11x+5\right)$

$Factor: 2x^2-11x+5$

$2x^2-11x+5=\left(2x^2-x\right)+\left(-10x+5\right)$

$=x\left(2x-1\right)-5\left(2x-1\right)$

$2x^3-5x^2-28x+15=\left(x+3\right)\left(x-5\right)\left(2x-1\right)$

$\left(x+3\right)\left(x-5\right)\left(2x-1\right)=0$

$\mathrm{Using\:the\:Zero\:Factor\:Principle:}$

thus zeros of f(x) is

$x=-3,\:x=5,\:x=\frac{1}{2}$

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